We offer you a convenient free online calculator for solving quadratic equations. You can quickly get and understand how they are solved using clear examples.
To produce solve quadratic equation online, first bring the equation to its general form:
ax 2 + bx + c = 0
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How to solve a quadratic equation
| How to solve a quadratic equation: | Types of roots: |
|
1.
Reduce the quadratic equation to its general form: General view Аx 2 +Bx+C=0 Example: 3x - 2x 2 +1=-1 Reduce to -2x 2 +3x+2=0 2.
Find the discriminant D. 3.
Finding the roots of the equation. |
1.
Real roots. Moreover. x1 is not equal to x2 The situation occurs when D>0 and A is not equal to 0. 2.
The real roots are the same. x1 equals x2 3.
Two complex roots. x1=d+ei, x2=d-ei, where i=-(1) 1/2 5.
The equation has countless solutions. 6.
The equation has no solutions. |
To consolidate the algorithm, here are a few more illustrative examples of solutions to quadratic equations.
Example 1. Solving an ordinary quadratic equation with different real roots.
x 2 + 3x -10 = 0
In this equation
A=1, B=3, C=-10
D=B 2 -4*A*C = 9-4*1*(-10) = 9+40 = 49
We will denote the square root as the number 1/2!
x1=(-B+D 1/2)/2A = (-3+7)/2 = 2
x2=(-B-D 1/2)/2A = (-3-7)/2 = -5
To check, let's substitute:
(x-2)*(x+5) = x2 -2x +5x – 10 = x2 + 3x -10
Example 2. Solving a quadratic equation with matching real roots.
x 2 – 8x + 16 = 0
A=1, B = -8, C=16
D = k 2 – AC = 16 – 16 = 0
X = -k/A = 4
Let's substitute
(x-4)*(x-4) = (x-4)2 = X 2 – 8x + 16
Example 3. Solving a quadratic equation with complex roots.
13x 2 – 4x + 1 = 0
A=1, B = -4, C=9
D = b 2 – 4AC = 16 – 4*13*1 = 16 - 52 = -36
The discriminant is negative – the roots are complex.
X1=(-B+D 1/2)/2A = (4+6i)/(2*13) = 2/13+3i/13
x2=(-B-D 1/2)/2A = (4-6i)/(2*13) = 2/13-3i/13
, where I is the square root of -1
Here are actually all the possible cases of solving quadratic equations.
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In this article we will learn to solve biquadratic equations.
So, what type of equations are called biquadratic?
All equations of the form ah 4 +
bx
2
+
c
= 0
, Where a ≠ 0, which are square with respect to x 2, and are called biquadratic equations. As you can see, this entry is very similar to the entry for a quadratic equation, so we will solve biquadratic equations using the formulas that we used to solve the quadratic equation.
Only we will need to introduce a new variable, that is, we denote x 2 another variable, for example at or t (or any other letter of the Latin alphabet).
For example, let's solve the equation x 4 + 4x 2 ‒ 5 = 0.
Let's denote x 2
through at
(x 2 = y
) and we get the equation y 2 + 4y – 5 = 0.
As you can see, you already know how to solve such equations.
We solve the resulting equation:
D = 4 2 – 4 (‒ 5) = 16 + 20 = 36, √D = √36 = 6.
y 1 = (‒ 4 – 6)/2= ‒ 10 /2 = ‒ 5,
y 2 = (‒ 4 + 6)/2= 2 /2 = 1.
Let's return to our variable x.
We found that x 2 = ‒ 5 and x 2 = 1.
We note that the first equation has no solutions, and the second gives two solutions: x 1 = 1 and x 2 = ‒1. Be careful not to lose the negative root (most often they get the answer x = 1, but this is not correct).
Answer:- 1 and 1.
To better understand the topic, let's look at a few examples.
Example 1. Solve the equation 2x 4 ‒ 5 x 2 + 3 = 0.
Let x 2 = y, then 2y 2 ‒ 5y + 3 = 0.
D = (‒ 5) 2 – 4 2 3 = 25 ‒ 24 = 1, √D = √1 = 1.
y 1 = (5 – 1)/(2 2) = 4 /4 =1, y 2 = (5 + 1)/(2 2) = 6 /4 =1.5.
Then x 2 = 1 and x 2 = 1.5.
We get x 1 = ‒1, x 2 = 1, x 3 = ‒ √1.5, x 4 = √1.5.
Answer: ‒1; 1; ‒ √1,5; √1,5.
Example 2. Solve the equation 2x 4 + 5 x 2 + 2 = 0.
2y 2 + 5y + 2 =0.
D = 5 2 – 4 2 2 = 25 ‒ 16 = 9, √D = √9 = 3.
y 1 = (‒ 5 – 3)/(2 2) = ‒ 8 /4 = ‒2, y 2 = (‒5 + 3)/(2 2) = ‒ 2 /4 = ‒ 0.5.
Then x 2 = - 2 and x 2 = - 0.5. Please note that none of these equations have a solution.
Answer: there are no solutions.
Incomplete biquadratic equations- it is when b = 0 (ax 4 + c = 0) or c = 0
(ax 4 + bx 2 = 0) are solved like incomplete quadratic equations.
Example 3. Solve the equation x 4 ‒ 25x 2 = 0
Let's factorize, put x 2 out of brackets and then x 2 (x 2 ‒ 25) = 0.
We get x 2 = 0 or x 2 ‒ 25 = 0, x 2 = 25.
Then we have roots 0; 5 and – 5.
Answer: 0; 5; – 5.
Example 4. Solve the equation 5x 4 ‒ 45 = 0.
x 2 = ‒ √9 (has no solutions)
x 2 = √9, x 1 = ‒ 3, x 2 = 3.
As you can see, if you can solve quadratic equations, you can also solve biquadratic equations.
If you still have questions, sign up for my lessons. Tutor Valentina Galinevskaya.
website, when copying material in full or in part, a link to the original source is required.
Goals:
- Systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
- Deepen your knowledge by completing a number of tasks, some of which are unfamiliar either in type or method of solution.
- Forming an interest in mathematics through the study of new chapters of mathematics, nurturing a graphic culture through the construction of graphs of equations.
Lesson type: combined.
Equipment: graphic projector.
Visibility: table "Viete's Theorem".
During the classes
1. Oral counting
a) What is the remainder of the division of the polynomial p n (x) = a n x n + a n-1 x n-1 + ... + a 1 x 1 + a 0 by the binomial x-a?
b) How many roots can a cubic equation have?
c) How do we solve equations of the third and fourth degrees?
d) If b is an even number in a quadratic equation, then what is the value of D and x 1; x 2
2. Independent work (in groups)
Write an equation if the roots are known (answers to tasks are coded) “Vieta’s Theorem” is used
1 group
Roots: x 1 = 1; x 2 = -2; x 3 = -3; x 4 = 6
Make up an equation:
B=1 -2-3+6=2; b=-2
c=-2-3+6+6-12-18= -23; c= -23
d=6-12+36-18=12; d= -12
e=1(-2)(-3)6=36
x 4 -2 x 3 - 23x 2 - 12 x + 36 = 0(this equation is then solved by group 2 on the board)
Solution . We look for whole roots among the divisors of the number 36.
р = ±1;±2;±3;±4;±6…
p 4 (1)=1-2-23-12+36=0 The number 1 satisfies the equation, therefore =1 is the root of the equation. According to Horner's scheme
p 3 (x) = x 3 - x 2 -24x -36
p 3 (-2) = -8 -4 +48 -36 = 0, x 2 = -2
p 2 (x) = x 2 -3x -18=0
x 3 =-3, x 4 =6
Answer: 1;-2;-3;6 sum of roots 2 (P)
2nd group
Roots: x 1 = -1; x 2 = x 3 =2; x 4 =5
Make up an equation:
B=-1+2+2+5-8; b= -8
c=2(-1)+4+10-2-5+10=15; c=15
D=-4-10+20-10= -4; d=4
e=2(-1)2*5=-20;e=-20
8+15+4x-20=0 (group 3 solves this equation on the board)
р = ±1;±2;±4;±5;±10;±20.
p 4 (1)=1-8+15+4-20=-8
р 4 (-1)=1+8+15-4-20=0
p 3 (x) = x 3 -9x 2 +24x -20
p 3 (2) = 8 -36+48 -20=0
p 2 (x) = x 2 -7x +10 = 0 x 1 = 2; x 2 =5
Answer: -1;2;2;5 sum of roots 8(P)
3 group
Roots: x 1 = -1; x 2 =1; x 3 = -2; x 4 =3
Make up an equation:
В=-1+1-2+3=1;В=-1
с=-1+2-3-2+3-6=-7;с=-7
D=2+6-3-6=-1; d=1
e=-1*1*(-2)*3=6
x 4 - x 3- 7x 2 + x + 6 = 0(group 4 solves this equation later on the board)
Solution. We look for whole roots among the divisors of the number 6.
р = ±1;±2;±3;±6
p 4 (1)=1-1-7+1+6=0
p 3 (x) = x 3 - 7x -6
р 3 (-1) = -1+7-6=0
p 2 (x) = x 2 - x -6 = 0; x 1 = -2; x 2 =3
Answer: -1;1;-2;3 Sum of roots 1(O)
4 group
Roots: x 1 = -2; x 2 = -2; x 3 = -3; x 4 = -3
Make up an equation:
B=-2-2-3+3=-4; b=4
c=4+6-6+6-6-9=-5; с=-5
D=-12+12+18+18=36; d=-36
e=-2*(-2)*(-3)*3=-36;e=-36
x 4 +4x 3 – 5x 2 – 36x -36 = 0(this equation is then solved by group 5 on the board)
Solution. We look for whole roots among the divisors of the number -36
р = ±1;±2;±3…
p(1)= 1 + 4-5-36-36 = -72
p 4 (-2) = 16 -32 -20 + 72 -36 = 0
p 3 (x) = x 3 +2x 2 -9x-18 = 0
p 3 (-2) = -8 + 8 + 18-18 = 0
p 2 (x) = x 2 -9 = 0; x=±3
Answer: -2; -2; -3; 3 Sum of roots-4 (F)
5 group
Roots: x 1 = -1; x 2 = -2; x 3 = -3; x 4 = -4
Write an equation
x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by group 6 on the board)
Solution . We look for whole roots among the divisors of the number 24.
р = ±1;±2;±3
p 4 (-1) = 1 -10 + 35 -50 + 24 = 0
p 3 (x) = x- 3 + 9x 2 + 26x+ 24 = 0
p 3 (-2) = -8 + 36-52 + 24 = O
p 2 (x) = x 2 + 7x+ 12 = 0
Answer: -1;-2;-3;-4 sum-10 (I)
6 group
Roots: x 1 = 1; x 2 = 1; x 3 = -3; x 4 = 8
Write an equation
B=1+1-3+8=7;b=-7
c=1 -3+8-3+8-24= -13
D=-3-24+8-24= -43; d=43
x 4 - 7x 3- 13x 2 + 43x - 24 = 0 (this equation is then solved by group 1 on the board)
Solution . We look for whole roots among the divisors of the number -24.
p 4 (1)=1-7-13+43-24=0
p 3 (1)=1-6-19+24=0
p 2 (x)= x 2 -5x - 24 = 0
x 3 =-3, x 4 =8
Answer: 1;1;-3;8 sum 7 (L)
3. Solving equations with a parameter
1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is equal to (-1)
Write the answer in ascending order
R=P 3 (-1)=-1+3-m-15=0
x 3 + 3x 2 -13x - 15 = 0; -1+3+13-15=0
By condition x 1 = - 1; D=1+15=16
P 2 (x) = x 2 +2x-15 = 0
x 2 = -1-4 = -5;
x 3 = -1 + 4 = 3;
Answer: - 1; -5; 3
In ascending order: -5;-1;3. (b N S)
2. Find all roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders from its division into binomials x-1 and x +2 are equal.
Solution: R=P 3 (1) = P 3 (-2)
P 3 (1) = 1-3 + a- 2a + 6 = 4-a
P 3 (-2) = -8-12-2a-2a + 6 = -14-4a
x 3 -Zx 2 -6x + 12 + 6 = x 3 -Zx 2 -6x + 18
x 2 (x-3)-6(x-3) = 0
(x-3)(x 2 -6) = 0
3) a=0, x 2 -0*x 2 +0 = 0; x 2 =0; x 4 =0
a=0; x=0; x=1
a>0; x=1; x=a ± √a
2. Write an equation
1 group. Roots: -4; -2; 1; 7;
2nd group. Roots: -3; -2; 1; 2;
3 group. Roots: -1; 2; 6; 10;
4 group. Roots: -3; 2; 2; 5;
5 group. Roots: -5; -2; 2; 4;
6 group. Roots: -8; -2; 6; 7.
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